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# From the consecutive integers -10 to 10 inclusive, 20 integers are ran

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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

The consecutive integers from -10 to 10 are: -10, -9, -8, -7 ... -1, 0, 1, ... 9, 10

You have to select 20 integers with repetition.

So you can select all 10. The product will be $$10^{20}$$.
or
You can select all -10. The product will be $$(-10)^{20}$$ (which is essentially same as $$10^{20}$$)
or
You can select ten 1s and ten -10s. The product will be $$(-10)^{10}$$
or
You can select 0 and any other 19 numbers. The product will be 0.
or
You can select -1 and nineteen 10s. The product will be $$-(10)^{19}$$.

But how will you get $$- (10)^{20}$$?

Editing here itself - I missed out the case in which you can get $$- (10)^{20}$$ (by multiplying odd number of -10 with rest of the 10s). Possibly because I mis-read the question.
'Least possible value of the product' to me implied the product that is not possible. Since all options a to d were possible, I overlooked that option (E) is possible too.

Actually the question should read 'the smallest possible value of the product' to avoid such confusion. Since -10^20 is the smallest value that can be obtained (greatest magnitude but negative), the answer is (E).

Originally posted by KarishmaB on 08 Jul 2012, 20:59.
Last edited by KarishmaB on 09 Jul 2012, 03:00, edited 1 time in total.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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Video solution from Quant Reasoning:
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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Bunuel wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$ .

I'm not quiet sure i understand, why choose only -10 and 10, there are other integers included in the set of -10 to 10. Also, isn't 10^1*10^(-19) = 10^(-18)?
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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catty2004 wrote:
Bunuel wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$ .

I'm not quiet sure i understand, why choose only -10 and 10, there are other integers included in the set of -10 to 10. Also, isn't 10^1*10^(-19) = 10^(-18)?

We choose -10 and 10 because this way we get the least product (notice that -10^(20) is the smallest number among answer choices).

Also it's $$10^1*(-10)^{19}=-10^{20}$$ not $$10^1*(10)^{-19}=-10^{20}$$.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
Thanks for pointing that out!!! Silly me. But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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VeritasPrepKarishma wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

The consecutive integers from -10 to 10 are: -10, -9, -8, -7 ... -1, 0, 1, ... 9, 10

You have to select 20 integers with repetition.

So you can select all 10. The product will be $$10^{20}$$.
or
You can select all -10. The product will be $$(-10)^{20}$$ (which is essentially same as $$10^{20}$$)
or
You can select ten 1s and ten -10s. The product will be $$(-10)^{10}$$
or
You can select 0 and any other 19 numbers. The product will be 0.
or
You can select -1 and nineteen 10s. The product will be $$-(10)^{19}$$.

But how will you get $$- (10)^{20}$$?
You need to select -1 and twenty 10s but you cannot select 21 numbers. You cannot have the product as negative $$10^{20}$$.
Hence (E) is not possible
.

E is possible if you select 10 odd number of times and -10 the remaining number of times. For example:

$$10^1*(-10)^{19}=-10^{20}$$ or $$10^3*(-10)^{17}=-10^{20}$$ or $$10^5*(-10)^{15}=-10^{20}$$... or $$10^{19}*(-10)^{1}=-10^{20}$$.

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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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catty2004 wrote:
But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

Actually, the question means "smallest possible value of the product"

$$-10^{20}$$ is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
VeritasPrepKarishma wrote:
catty2004 wrote:
But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

Actually, the question means "smallest possible value of the product"

$$-10^{20}$$ is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.

Yes, the question means "what is the smallest possible value of the product of the 20 integers?"
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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i missed out the bracket for -10^20. I was assuming that value to be positive as a negative number raised to positive power is positive. Nice problem![/edit]
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Bigred2008 wrote:
From the consecutive integers -10 to 10, inclusive 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of 20 integers.

A. (-10)^20
B. (-10)^10
C. 0
D. -(10)^19
E. -(10)^20

The smallest possible value of the product will be negative (since it has to be smallest) with highest absolute value i.e. the number should be as left on the number line as possible.
Out of all the options, (E) is the smallest. Let's see if it is possible.
If you select all 10s and only one -10, you will get $$-10*(10)^{19}$$ which is the same as $$-(10)^{20}$$. So it is possible to get the product given in (E).

Notice that options (A) and (B) are positive.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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Hi,

I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.

So what I though it that we need an odd number of negative values to have a negative value.
To have the greatest negative value we need -10 (the higest possible value), the greatest number of odd times; so in out case 19.
We still have to use one number, so 10 would be the final choise, as it would make the negative 19(-10) 10 times greater.

However, for me it is:
-10(19) * 10 (1) instead of 10^1 * -10^19 = -10^20.

I understand the 10^1, but -10(19) is not -10^19. This is where I am lost...

So, why the powers?
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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pacifist85 wrote:
Hi,

I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.

So what I though it that we need an odd number of negative values to have a negative value.
To have the greatest negative value we need -10 (the higest possible value), the greatest number of odd times; so in out case 19.
We still have to use one number, so 10 would be the final choise, as it would make the negative 19(-10) 10 times greater.

However, for me it is:
-10(19) * 10 (1) instead of 10^1 * -10^19 = -10^20.

I understand the 10^1, but -10(19) is not -10^19. This is where I am lost...

So, why the powers?

You are given that you need to find the product of all 20 numbers. When you multiply -10 by -10, you get $$(-10)^2$$, not $$-10*2$$ (which is -10 + (-10))

Just like you multiplied the last 10 you picked, you have to multiply all previous 10s too.
So multiply -10, 19 times to get $$-10*-10*-10...*-10 = (-10)^{19}$$
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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mbaspire wrote:
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A $$(-10)^{20}$$
B $$(-10)^{10}$$
C 0
D $$-(10)^{19}$$
E $$-(10)^{20}$$

Choose nineteen -10's and one 10
So, the product = [(-10)^19][10]
Notice that [(-10)^19] is NEGATIVE, which means [(-10)^19][10] is also NEGATIVE.

So, [(-10)^19][10] = -[(10)^19][10]
= -(10)^20
= E

Cheers,
Brent
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
Bunuel wrote:
mbaspire wrote:
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A $$(-10)^{20}$$
B $$(-10)^{10}$$
C 0
D $$-(10)^{19}$$
E $$-(10)^{20}$$

Hi Bunuel,

I'm confused. I have referred to gmat club math book about multiply 2 number with exponents. the base should be the same to allow sum of the exponent while in this question the base is 10 & -10. So we can't sum 10^10 * (-10)^19. Do I miss something?

Thanks
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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Mo2men wrote:
Bunuel wrote:
mbaspire wrote:
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A $$(-10)^{20}$$
B $$(-10)^{10}$$
C 0
D $$-(10)^{19}$$
E $$-(10)^{20}$$

Hi Bunuel,

I'm confused. I have referred to gmat club math book about multiply 2 number with exponents. the base should be the same to allow sum of the exponent while in this question the base is 10 & -10. So we can't sum 10^10 * (-10)^19. Do I miss something?

Thanks

Notice that (-10)^19 = (-1)*(10)^19. This way you'll get the same base.

Hope it's clear.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
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catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

This problem is testing our knowledge of the multiplication rules for positive and negative numbers. Remember that when we multiply an even number of negative numbers together the result is positive and when we multiply an odd number of negative numbers together the result is negative.

Because we are selecting 20 numbers from the list, we want to start by selecting the smallest 19 numbers we can and multiplying those together. In our list the smallest number we can select is -10. So we have:

(-10)^19 (Note that this product will be negative.)

Since we need to select a total of 20 numbers we must select one additional number from the list. However, since the final product must be as small as possible, we want the final number we select to be the largest positive value in our list. The largest positive value in our list is 10. So the product of our 20 integers is:

(-10)^19 x 10 (Note that this product will still be negative.)

This does not look identical to any of our answer choices. However, notice that
(-10)^19 can be rewritten as -(10)^19, so

(-10)^19 x 10 = -(10)^19 x (10)^1 = -(10)^20.

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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran [#permalink]
Bunuel wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$.